Question

In $△$ ABC, $D$ is a point on $BC$ such that $AB=AD=BD=DC$, then:
$\angle ADC:\angle C=3:1$.

• A. True
• B. False

Answer 1

The correct option is B False

Here, $AB,BD,AD$ are sides of a triangle. These sides are given equal. Hence the triangle is an equilateral triangle.
Thus, $\angle ADB={60}^{\circ }$
In $△ADC$,
$AD=CD$
Thus, $\angle ADC=\angle ACD=y$ (Isosceles triangle property)
Now, $\angle ADB=\angle ADC+\angle ACD$
$60=y+y$
$y={30}^{\circ }$
$\angle ACD={30}^{\circ }$
$\angle ADC+\angle ADB=180$ (Angles on a straight line)
$60+\angle ADC=180$
$\angle ADC={120}^{\circ }$
Hence, $\frac{\angle ADC}{\angle C}=\frac{120}{30}$
$\frac{\angle ADC}{\angle C}=4:1$